# Power Series and Analyticity of a complex function

I was studying about the connection of analytic function and their power series representation.

Finally, I came to an understanding that, if I am given with an function, analytic at some point 'a', then I will be able to write a power series representation of that function, where that power series representation is convergent in some circle centered around that 'a'. Now, what about the behavior points outside this circle of convergence? Can the function remain analytic at those points?

In short, is it true if a function having a power series representation about a point is not convergent at a point outside the radius of convergence, then we cannot say about the analyticity of that function at that point.

Is my understanding correct? Or Am I still missing the essence of the power series expansion?

• We can't even say anything about analyticity within the disc of convergence. For instance, the power series representation of $\operatorname{Log}(z)$ centered at $z_0$ has radius of convergence $\vert z_0\vert$. But if we take $z_0=i-1$, then $\operatorname{Log}(z)$ isn't even defined on the entire disc of convergence, since the disc contains parts of the negative real axis, on which the main branch of the logarithm isn't defined. Jul 18, 2020 at 11:21
• Got it. But if some function is analytic at $z_0$ then it has to be analytic within that disc of convergence. And in the exterior it may or may not be analytic. Am I right?
– MB17
Jul 18, 2020 at 15:04
• No. The radius of convergence doesn't tell you anything about wether the function actually agrees with the power series on the entire disc of convergence. Analyticity only tells you that there is some neighborhood of $z_0$ in which the function agrees with a power series. There's nothing that would guarantee that this neighborhood is equal to the disc of convergence. Jul 18, 2020 at 15:39
• Isnt it true that the radius of convergence of an analytic function at $z_0$ is the distance of $z_0$ from its nearest singularity? And this does suggest that the function is analytic in that circle of convergence.
– MB17
Jul 18, 2020 at 16:39
• No, this is not true in general. A power series may define an analytic function on the disc of convergence, but there may be other functions which agree with the power series on a small neighborhood of $z_0$, but not on the entire disc of convergence. In such a case we can't say anything about singularities except that they are not in this small neighborhood. Jul 19, 2020 at 10:01

Your understanding is correct. Suppose that you define$$\begin{array}{rccc}f\colon&\Bbb C&\longrightarrow&\Bbb C\\&z&\mapsto&\begin{cases}\frac1{1-z}&\text{ if }|z|<1\\0&\text{ otherwise.}\end{cases}\end{array}$$Then, at $$D(0,1)$$, you have$$f(z)=1+z+z^2+\cdots$$and the radius of convergence of the series $$1+z+z^2+\cdots$$ is $$1$$. But $$f$$ is not analytic. It's not even continuous.

• Thank you. And is it true that the function, you defined is analytic in the disc $|z|<1$, and nowhere analytic outside the disc? And can we also define some functions that is analytic even outside the circle of convergence?
– MB17
Jul 18, 2020 at 15:08
• The function $f$ is analytic outside $D(0,1)$. For instance, at $D(2,1)$ you have$$f(z)=0+0\times(z-2)+0\times(z-2)^2+\cdots.$$ Jul 18, 2020 at 15:48
• Oh. Right. I didnt see through it. So there exists a power series expansion around every point except at its singularity. Thank you.
– MB17
Jul 18, 2020 at 16:42
• Not quite. If $w\in\Bbb C$, then you have $f(z)=\sum_{n=0}^\infty a_n(z-w)$ for some power series $\sum_{n=0}^\infty a_n(z-w)$ on some disk centered at $w$ if and only if $|w|\ne1$. Jul 18, 2020 at 16:46
• Got it. And just one more doubt, this is reference to previous comments in my question. That is, speaking of general sense, is a function always analytic inside the circle of convegrence?
– MB17
Jul 18, 2020 at 16:54